Question

If heat at constant volume for combustion of CH4 at 127°C is -20 kJ, then heat at constant pressure will be what ?

​

Answer 1

Answer:

The correct option is B)

Combustion of benzene:

C

6

​

H

6

​

(l)+

2

15

​

O

2

​

(g)⟶6CO

2

​

(g)+3H

2

​

O(l)

ΔH=ΔU+PΔV;                         Δn=6−

2

15

​

=−1.5

ΔH−ΔU=PΔV

ΔH−ΔU=ΔnRT

ΔH−ΔU=−1.5×R×(127+273)

ΔH−ΔU=−600R.

Explanation:

Answer 2

Answer:

-26.7kJ heat is produced.

Explanation:

The combustion of $CH_{4}$ is given as following:

$CH_{4} (g) + 2O_{2} (g)$ → $CO_{2} (g) + 2H_{2}O (l)$

We find the difference between the number of gaseous molecules of products and the number of gaseous molecule of reactants in the reaction.

Δ$n_{g}$ = $n_{P} - n_{R}$

Δ $n_{g}$ = 1 - (1 + 2)

Δ$n_{g}$ = -2

We use the enthalpy equation below to find the heat.

Δ$H$ = Δ$U +$ Δ$n_{g}RT$

We have the following data:

1. Internal Energy (U) = -20kJ = -20000J
2. Temperature = 127°C = (127 + 273.15 = 400.15K)
3. Δ$n_{g}$ = -2

ΔH = (-20000) + (-2)(8.31)(400.15)

ΔH = -20000 + (-6650.49)

ΔH = -26650.49 J ≈ -26700 J OR -26.7kJ